Given the equation: $ y = -4x^2 + 32x - 59$ Find the parabola's vertex.
Solution: When the equation is rewritten in vertex form like this, the vertex is the point $({h}, {k})$ $ y = A(x - {h})^2 + {k} $ We can rewrite the equation in vertex form by completing the square. First, move the constant term to the left side of the equation: $ \begin{eqnarray} y &=& -4x^2 + 32x - 59 \\ \\ y + 59 &=& -4x^2 + 32x \end{eqnarray} $ Next, we can factor out a $-4$ from the right side: $ y + 59 = -4(x^2 - 8x) $ We can complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $-8$ , so half of it would be $-4$ , and squaring that gives us ${16}$ . Because we're adding the $16$ inside the parentheses on the right where it's being multiplied by $-4$ , we need to add ${-64}$ to the left side to make sure we're adding the same thing to both sides. $ \begin{eqnarray} y + 59 &=& -4(x^2 - 8x) \\ \\ y + 59 + {-64} &=& -4(x^2 - 8x + {16}) \\ \\ y - 5 &=& -4(x^2 - 8x + 16) \end{eqnarray} $ Now we can rewrite the expression in parentheses as a squared term: $ y - 5 = -4(x - 4)^2 $ Move the constant term to the right side of the equation. Now the equation is in vertex form: $ y = -4(x - 4)^2 + 5 $ Now that the equation is written in vertex form, the vertex is the point $({h}, {k})$ $ y = A(x - {h})^2 + {k} $ $ y = -4(x - {(4)})^2 + {(5)} $ The vertex is $({4}, {5})$. Be sure to pay attention to the signs when interpreting an equation in vertex form.